Question: The lifespans of meerkats in a particular zoo are normally distributed. The average meerkat lives $12.7$ years; the standard deviation is $1.5$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a meerkat living less than $15.7$ years.
Solution: $12.7$ $11.2$ $14.2$ $9.7$ $15.7$ $8.2$ $17.2$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $12.7$ years. We know the standard deviation is $1.5$ years, so one standard deviation below the mean is $11.2$ years and one standard deviation above the mean is $14.2$ years. Two standard deviations below the mean is $9.7$ years and two standard deviations above the mean is $15.7$ years. Three standard deviations below the mean is $8.2$ years and three standard deviations above the mean is $17.2$ years. We are interested in the probability of a meerkat living less than $15.7$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the meerkats will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the meerkats will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $9.7$ years and the other half $({2.5\%})$ will live longer than $15.7$ years. The probability of a particular meerkat living less than $15.7$ years is ${95\%} + {2.5\%}$, or $97.5\%$.